Question

graph each equation.

1. \\(\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1\\)

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, and origin at (0,0)

Explanation:

Step1: Identify the conic section

The equation \(\frac{x^2}{4}+\frac{y^2}{9} = 1\) is in the standard form of an ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (since \(a^2 = 9\) and \(b^2=4\), and \(a>b\), so it is a vertical ellipse).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\), the center is at \((0,0)\) (since there are no \(h\) and \(k\) shifts, \(h = 0,k = 0\)).

• The length of the semi - major axis \(a=\sqrt{9}=3\), so the vertices are at \((0,\pm a)=(0, 3)\) and \((0,- 3)\).
• The length of the semi - minor axis \(b=\sqrt{4} = 2\), so the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

Plot the center \((0,0)\), vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((- 2,0)\). Then draw a smooth curve connecting these points to form the ellipse.

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\). The ellipse is drawn by connecting these points smoothly. (The actual graphing is done by plotting the key points and drawing the ellipse as described above on the given coordinate grid.)